How To Optimize Traffic

Transcription

1 Adaptive Threshold Policies for Multi-Channel Call Centers Benjamin Legros a Oualid Jouini a Ger Koole b a Laboratoire Génie Industriel, Ecole Centrale Paris, Grande Voie des Vignes, Châtenay-Malabry, France b Department of Mathematics, VU University Amsterdam, De Boelelaan 1081a, 1081 HV Amsterdam, The Netherlands benjamin.legros@centraliens.net oualid.jouini@ecp.fr ger.koole@vu.nl IIE Transactions. To appear, Abstract In the context of multi-channel call centers with inbound calls and s, we consider a threshold policy on the reservation of agents for the inbound calls. We study a general nonstationary model where calls arrive according to a non-homogeneous Poisson process. The optimization problem consists of maximizing the throughput of s under a constraint on the waiting time of inbound calls. We propose an efficient adaptive threshold policy that is easy to implement in the Automatic Call Distributor ACD). This scheduling policy is evaluated through a comparison with the optimal performance measures found in the case of a constant arrival rate, and also a comparison with other intuitive adaptive threshold policies in the general non-stationary case. 1 Introduction Call centers require a very precise match of demand and supply. The delay of the treatment of a call, its waiting time, is usually not allowed to exceed 20 seconds Koole 2013)). Thus a very accurate prediction of the demand is required. However, this can rarely be obtained, because of the volatility of call arrival patterns. Therefore there is often a mismatch between demand and the scheduled supply, consisting of rostered call center employees usually called agents). Moreover, even if the demand is accurately forecasted, a considerable overcapacity should be scheduled to be able to deal with the random Poisson fluctuations of the demand. Usually queueing models are used to quantify this overcapacity, most often Erlang C. To prevent idle overcapacity, and to limit the necessity to have extremely accurate forecasts, inbound calls are sometimes mixed with other types of customer contacts which have a less strict delay requirements, such as s or outbound calls. This is called call) blending. The amount

2 of capacity assigned to the other channels is supposed to adapt to the number of inbound calls, giving at the same time a good service level for the inbound calls and a good occupancy of the call center agents. Because of the strict waiting time requirement on inbound calls it is best to give them priority over the other channels. To maximize agent productivity it would be optimal to assign an outbound job to every idle agent when there are no inbound calls in queue. This would lead to a 100% productivity. However, this policy leads to long waiting times for inbound calls because never an agent is waiting for an inbound call to arrive. Consequently, the service level constraint on inbound calls might be violated. For instance, consider a simple Markovian queueing model with inbound jobs arriving at rate λ arriving at a queue with infinite capacity, an infinite amount of s, s polyvalent identical agents, identical service rate µ for both job types. Using the underlying birthdeath process, one may easily deduce that the expected waiting time for inbound jobs is ρ λ1 ρ), where ρ = λ sµ, and the probability of delay of inbound jobs is 1. Therefore, for a given staffing level, this work-conserving policy has not enough flexibility to reach some predefined service level for inbound jobs. Somehow we should reserve capacity for inbound jobs to obtain an expected waiting time strictly lower than ρ λ1 ρ) and a probability of delay strictly lower than 1, which allows to reach better call service levels. Thus a more sophisticated assignment policy, other than the work-conserving one, is required. A service level measures a call waiting time performance for example the proportion of calls that are answered within a predefined time threshold, or the expected waiting time). In Bhulai and Koole 2003) and Gans and Zhou 2003) it is shown that the optimal assignment policy is of the following form: outbound jobs should only be scheduled when there are no waiting inbound calls and when the number of idle agents exceeds a certain threshold. Thus the problem of controlling our blended call center reduces to determining the right threshold level. This threshold however depends on all the system parameters, which are the inbound call arrival rate, the inbound call service rate, the service rate and the number of agents. But these parameters, especially the arrival rate, are often hard to determine. This calls for a policy in which the threshold is adapted to the current situation without using explicitly the parameters of the system. In this paper such adaptive policies are studied, both for systems with a constant but unknown) arrival rate and for the more realistic situation of a fluctuating arrival rate. The parameter that is used to update the threshold is the service level up to that moment, a number which is always available in call centers. We consider the service level that measures the proportion of calls that are answered within a predefined time threshold. The overall objective is to reach a certain service level by the end of the day, while maximizing the number of s that are done. 2

3 We discuss the relevant literature. There is a rich literature on planning and scheduling in call centers, see Gans et al. 2003); Akşin et al. 2007). However, few papers focus on blending. The general context of multi-channel call centers is described in Koole 2013), Chapter 7. Deslauriers et al. 2007) extend the earlier mentioned papers by having different types of agents. Outbound jobs are served only by multi-channel blended) agents, whereas inbound calls can be served by either inbound-only or blended agents. Inbound callers may balk or abandon. They evaluate several performance measures of interest, including the rate of outbound jobs and the proportion of inbound calls waiting more than some fixed number of seconds. A collection of continuous-time Markov chain CTMC) models which capture many real-world characteristics while maintaining parsimony that results in fast computation are presented. They discuss and explore the tradeoffs between model fidelity and efficacy and compare different CTMC models with a realistic simulation model of a Bell Canada call center. Armony and Ward 2010) present an optimization problem; the objective is to minimize steadystate expected customer waiting time subject to a fairness constraint on the workload division. They show that in such a problem, which is close to ours, a threshold policy outperforms a common routing policy used in call centers that routes to the agent that has been idle the longest). Milner and Olsen 2008) consider a call center with contract and non-contract customers. They explore the common use to give priority to contract customers only in off peaks. They show that this choice is a good one under classical assumptions such as stationarity). They also present examples when it is not. This result is important since we found an insight arguing that the service level for inbound calls has to be very strictly respected during off peaks. This paper is organized as follows. Section 2 presents our model. Sections 3 and 4 contain our results, first for a constant arrival rate in Section 3 and then in Section 4 with a fluctuating arrival rate. We end with a short conclusion. 2 Model We consider a call center modeled as a multi-server queueing system with two types of jobs, foreground jobs inbound calls) and background jobs s). The arrival process of calls is assumed to be a non-homogeneous Poisson process with rate λt), for t 0. Calls arrive at a dedicated first come, first served FCFS) queue with infinite capacity. There is an infinite supply of background jobs, waiting for treatment in a dedicated FCFS queue. There are s identical, parallel servers agents in call center parlance). Each agent can handle both types of jobs. We assume that the service times of foreground and background jobs are exponentially distributed with rates µ and µ 0, 3

4 respectively. Neither abandonment nor retrials are modeled. Foreground jobs are more important than background ones in the sense that the former request a quasi-instantaneous answer waiting time in the order of seconds or minutes), while the latter are more flexible and could be delayed for several tens of) hours. The objective of the call center manager over a working day is to maximize the throughput while satisfying a constraint on the call waiting time in the queue. Since the model is transient, we can not define the waiting time of an arbitrary customer as a unique random variable. There is a random number of served customers during the working period, say Q. If Q > 0, we denote by W n the random variable for the waiting time of customer n, for n {1,..., Q}. We want that the expected proportion of ) calls that wait less than a predefined threshold τ is at least equals to α, i.e., E Q 1 Q α, for τ 0 and 0 α 1. Note 1 Wn τ n=1 that we do not consider arriving customers at the end of the working period which can not be served. We then aim to find the best routing rules in terms of efficiency for the considered problem and easiness of implementation in call center software. We assume that preemption of jobs in service is not allowed. This is a quite natural assumption. An agent usually prefers to finish answering an underway outbound job rather than starting it over later on. This is also preferred from an efficiency perspective. Evidently, when the background jobs are outbound calls, then it is neither acceptable to preempt. For a similar model but with a constant arrival rate and equal service requirements for the two job types, Bhulai and Koole 2003) prove that the optimal policy is a threshold policy with the priority given to calls some servers reserved for calls). Their result is mainly based on the fact that it is optimal to handle calls as long as the queue of calls is not empty. For our general modeling, the analysis is more complicated. Even for a constant arrival rate but different service requirements, the optimal policy is hard to obtain, and might not be useful in practice for software implementation for example). For simplicity and usefulness of the results in practice, we then restrict ourselves to the case of threshold policies. Moreover, Bhulai and Koole 2003) numerically show, for more general cases, that the appealing threshold policies are good approximations of the optimal ones. More concretely, the functioning of the call center under a threshold policy is as follows. Let us denote the threshold by u, 0 u s. Upon arrival, a call is immediately handled by an available agent, if any. If not, the call waits in the queue. When an agent becomes idle, she handles the call at the head of the queue with calls, if any. If not, the agent may either handle an , or she remains idle. If the number of idle agents excluding her) is at least s u, then the agent in 4

5 question handles an . Otherwise, she remains idle. In other words, there are s u agents that are reserved for calls, so, there are at least u agents working at any time. In this paper, we propose an adaptive threshold policy which adjusts the threshold as a function of the process of the call service level. We divide the working day into N identical intervals, each with length θ. The total working duration in a day is D, D = Nθ. At the beginning of each interval i i = 1,..., N), we define the threshold u i, 0 u i s, under which the job routing policy works during interval i. Let T denote the expected throughput of s over the whole day, i.e., the ratio between the number of treated s and D. Let also SL be) the proportion, for the whole day, of calls that have waited less than τ, SL = E Q 1 Q, where Q is the random variable 1 Wn τ n=1 measuring the number of served customers during the whole day. In summary, our optimization problem can be formulated as { Maximize T subject to SL α, where the decision variables are u i with 0 u i s, for i = 1,..., N. It is clear that the best case for calls is such that u i = 0 for all i, which means that no is treated and SL is maximized case of an Mt)/M/s with only calls). We therefore assume from now on that the parameters λt) for t 0, µ and s are such that SL α for u i = 0 i = 1,..., N). 1) 3 Constant Arrival Rate We consider a basic case with a constant arrival rate, λt) = λ for t 0 and a constant threshold, u i = u for i = 1,..., N and 0 u s. The purpose of the analysis in this section is to understand the behavior of the performance measures as a function of the threshold in order to build an efficient method for the threshold adaptation rule u i for i = 1,..., N) in the case of a non-constant arrival rate. In Section 3.1 we propose a method to compute the performance measures, then in Section 3.2 we use them to provide a useful insight to construct our adaptive policy. 3.1 Performance Measures In Section we provide close form formula of the performance measures in the case of equal service rates and study the form of these measures as a function of the threshold. Then in Section we propose a numerical method to compute the performance measures in the case of unequal service rates. Since we consider a stationary model we can define a unique random variable for the waiting time of an arbitrary customer W, and denote by P W < τ) the probability that an 5

6 arbitrary customer waits less than τ τ > 0) Equal Service Rates We consider the case µ = µ 0. First, we compute the performance measures of interest for calls and s for a given constant reservation threshold, denoted by u, 0 u s. We then develop some structural results that will be used in Section 3.2. Let us define the stochastic process {xt), t 0}, where xt) {u, u + 1, u + 2, } is the number of jobs in service plus the number of jobs in the queue of calls. Since µ = µ 0, we need not distinguish between the two job types in service. The process {xt), t 0} is a birth-death process. It is similar to that of an M/M/s queue without the states {0, 1,, u 1}. The transition rate from state x to state x 1 is min{x, s}µ, for x > u, and that from state x to state x + 1 is λ, for x u. We denote by a the ratio λ λ µ. Also, under the stability condition sµ < 1, we denote by p x the steady-state probability to be in state x N. In Theorem 1, we give the expression of the throughput, T s, u, a), and that of the probability that the call waiting time is less than τ, SL = P W < τ). Theorem 1 For 0 u s, we have s u ) 1 ) a k u! T s, u, a) = µ u + k)! + as u u! a s u a k u! u + s! s a u + k 1)! + as u+1 u! λ, 2) s 1)!s a) k=1 P W < τ) = 1 Cs, u, a)e τsµ λ), 3) with Cs, u, a) = as u u! s u a k u! s!1 a/s) u + k)! + as u u! s! ) 1 a. 4) s a Proof. For 0 x < u, we have p x = 0. For 0 k s u, we have p u+k = ak u! u+k)! p u. For k 0, we have p s+k = ak s k p s. Since all probabilities sum up to one, we obtain s u ) 1 a k u! p u = u + k)! + as u u! a. 5) s! s a The throughput can be seen as the overall throughput of calls and s) minus the call 6

7 throughput. Thus After some algebra, we deduce that s u T s, u, a) = u + k)µp u+k + sµ p s+k λ. k=1 ) s u a T s, u, a) = µp u u k u! + u + k 1)! + as u+1 u! λ. s 1)!s a) k=1 Note that the lower bound of T s, u, a) is T s, 0, a) = 0, which corresponds to the case when all servers are reserved to calls. As for the upper bound, it is T s, s, a) = sµ λ, which corresponds to the case of no server reservation for calls the infinite amount of s leads to sµ as a total throughput for the two job types). The call service level, P W > τ), is obtained using the PASTA property. We have P W > τ) = p s+n P W > τ x = n + s), where P W > τ x = s + n) is the conditional probability that the n=0 waiting time of a new call exceeds τ, given that it finds all servers busy and n calls waiting ahead in the queue, n 0. It is easy to see that this conditional waiting time follows an Erlang distribution with n + 1 stages and a rate of sµ per stage. Then, P W > τ x = s + n) = leads to Observing that n=k a n n sµτ sµτ)k P W > τ) = p s s n e k! n=0 ) n = lim p s e sµτ sµτ) k a ) n. n k! s a ) n s = a ) k 1 s 1 a/s implies n=k n e sµτ sµτ) k k!, which P W > τ) = Cs, u, a)e τsµ λ), 6) with Cs, u, a) = p s 1 a/s = as u u! s u a k u! s!1 a/s) u + k)! + as u u! s! ) 1 a. 7) s a Note that the upper bound of Cs, u, a) is Cs, s, a) = 1 no server reservation for calls, then, any arriving call has to wait for service), and its lower bound is Cs, 0, a) = s ) 1 a k a s s!1 a/s) k! + as+1 s!s a) all servers are reserved to calls, which corresponds for calls to a standard M/M/s queue with no 7

8 s). This finishes the proof of the theorem. In Proposition 1, we prove monotonicity results of the system performance measures as a function of the threshold. Proposition 1 For a > 0, the following holds: 1. The throughput T is strictly increasing and neither convex nor concave in u, for 0 u s. However the end of the throughput, for 0 s 2 u s, is concave in u. 2. The call service level P W < τ) is strictly decreasing and concave in u, for 0 u s. Proof. Let us prove the first statement. From Equation 2), we have T s, u, a) = µ 1 u 1)! + a u! + 1 u! + a u+1)! + ) for 0 u s. Thus T s, u, a) = µ a + 1 g u λ, with g u = 1 u + a uu + 1) + + a2 u+1)! + + as u s 1)! + a2 u+2)! + + as u 1 s 1)! + a s u 1 uu + 1)u + 2) s 1) + for 0 < u s and T s, 0, a) = 0). We may write for 0 < u < s g u+1 g u = + 1 u ) + u a u + 1)u + 2) a uu + 1) a s u 1 u + 1)u + 2) s 1)s a) a s u 1 uu + 1) s 1) as u+1 s 1)!s a) as u s 1)!s a) λ, a s u uu + 1)u + 2) s 1)s a), ) + 8) ) + a s u uu + 1) s 1)s a) ). Since each term of the summation in the right hand of Equation 8) is strictly negative, g u+1 < g u for 0 < u < s. Then, g u is strictly decreasing in u for 0 < u s. We also have T s, 1, a) > 0 = T s, 0, a). This implies that T s, u, a) is strictly increasing in u, for 0 u s. Figure 1 illustrates that in general the throughput is neither convex nor concave. Let us now prove that the end of the throughput, for s 2 u s and a > 0, is concave in u. For s 2, we have T s, s, a) = sµ λ, T s, s 1, a) = µ s s2 s+a) λ, and T s, s 2, a) = µ s 2 s+a s3 3s 2 +2a+1)s 2a+a 2 ) λ. This implies T s, s 1, a) T s, s 2, a) = µs 1) ss 2 s+a) s2 a 2 ), and T s, s, a) T s, s 1, a) = µ s s a), which yields to T s, s 1, a) T s, s 2, a) = T s, s, a) T s, s 1, a)) s 1)s+a). Since for s 2 s 2 s+a and a > 0, s 1)s + a) s 2 s + a) = as 2) 0, we may write T s, s 1, a) T s, s 2, a) T s, s, a) T s, s 1, a). Then the end of the throughput is concave, which finishes the proof of the first statement of the proposition. 8

9 d d d ed dd d d e e dd Figure 1: throughput s = 10, µ 0 = µ = 0.2, λ = 1.4) In what follows, we prove the second statement of the proposition. Let us define the sequence f u as f u = s!1 a/s)cs, u, a), for 0 u s. Using Equation 3), it suffices then to prove that f u is strictly increasing and convex in u. We start by proving that f u is strictly increasing in u. We have for 0 u s. Since for 0 u < s we have s u a u s s u a k u+k)! k=1 ) s u 1 a f u = s!s a) + a k+u s, u + k)! ), we deduce that fu+1 1 f u 1 a k+u s u+k)! = a u s s u a k u+k)! 0 u < s. Then, P W < τ) is strictly decreasing in u, for 0 u s. ) and s u+1) a k+u+1 s u+1+k)! = = au s u! < 0. This implies that f u < f u+1, for We next focus on the proof of convexity of f u in u for s 2). We do so by proving that f u + f u+2 2f u+1 > 0, for 0 u s 2. Since f u +f u+2 2f u+1 = f u f u+1 f u+2 f 1 u+2 f u f u+1 1 f u 1 2fu+2 1 f u 1 ), it suffices to prove that fu+2 1 f u f u+1 1 f u 1 2fu+2 1 f u 1 > 0, for 0 u s 2. Observing that fu+1 1 = f u 1 au s u! and fu+2 1 = f u 1 au s u! au+1 s u+1)!, we obtain fu+2 1 f 1 = u+1 + f 1 f 1 u = au s u! u+1 f u 1 2f 1 u ) au s au+1 s u! fu u+2 f 1 f 1 u u + 1)! a ) + au s u + 1 u! au s ) + u! 1 + a u + 1 f 1 u )), ) ) au s fu 1 2 fu 1 au s au+1 s fu 1 u! u! u + 1)! for 0 u s 2. Since au s u! > 0, we thus need to prove that fu a ) + au s 1 + a ) > 0, u + 1 u! u + 1 9

10 for 0 u s 2. Using Equation 5), we may write fu 1 = au s u! p 1 u, for 0 u s 2. Therefore fu a ) + au s 1 + a ) = au s p 1 u 1 + u + 1 u! u + 1 u! for 0 u s 2. It remains then to prove that for 0 u s 2. p 1 u 1 + a ) a u + 1 u + 1 > 0, For 0 u s 2 and 0 < a < s, the derivative in a of p 1 u.) is given by a ) a ), u + 1 u + 1 p 1 u a ka k 1 u! u + k)! + u! s u + 1)a s u s a) + a s u+1 s! s a) 2 = s u s u ka k 1 u! = u + k)! + u! s! a s u s u + 1)s a) + a) s a) 2. Since 0 < a < s and u < s, we obtain p 1 u consequence, the expression p 1 u and 0 < a < s. Moreover, we have lim a 0, a > a a u+1 > 0. Thus p 1 u ) a u+1 p 1 u = lim a 0, a > 0 is strictly increasing in a, and as a is strictly increasing in a, for 0 u s 2 s u ) a k u! u + k)! + as u u! a. s! s a Since lim a 0, a > 0 a s u u! s! a s a = 0, and lim a 0, a > 0 s u a k u! u+k)! = 1, we get lim a 0, a > 0 p 1 u = 1. This implies lim a 0, a > 0 p 1 u 1 + a ) a u + 1 u + 1 = 0, ) for 0 u s 2. Using now the fact that p 1 u 1 + a u a u+1 is strictly increasing in a, ) we deduce that p 1 u 1 + a u a u+1 > 0, for a > 0 and 0 u s 2. This completes the proof of the proposition. 10

11 Note that in the particular remaining case a = 0 i.e., no calls, λ = 0), the throughput T is increasing and linear in u. We have T = uµ, for 0 u s Unequal Service Rates In this section we focus on the performance evaluation throughput and call waiting time distribution) for the case of unequal service rates, µ µ 0. In contrast to the case of equal service rates, the performance expressions are here too cumbersome to allow the development of useful structural results. The results of this section are however still useful for the numerical experiments in Section 3.2 in order to build the insights on the threshold policy for the more general case with a non-constant call arrival rate. As in Bhulai and Koole 2003), our approach consists in using a Markov chain analysis to derive the steady-state probabilities of the system, from which the performance measures are characterized thereafter. To simplify the presentation, we focus on the particular case u = s. The analysis for the case u = 0 is obvious, and that of the remaining cases, 0 < u < s, is done similarly to the case u = s. It simply adds a finite number of additional equations but does not impact the general form of the steady-state probabilities. Consider the stochastic process {xt), yt)), t 0}, where xt) is the number of waiting calls in the queue and yt) is the number of s being in service, x N, y {0, 1,, s}. This process is a Markov chain. For x 0 and 0 y s, the transition rate from x, y) to x + 1, y) is λ. For x 1 and 0 y s, the transition rate from x, y) to x 1, y) is s y)µ. For x 1 and 1 y s the transition rate from x, y) to x 1, y 1) is yµ 0. For 0 y s, the transition rate from 0, y) to 0, y + 1) is s y)µ. Because of the priority of inbound calls over s, no transition exists from x, y) to x, y 1), for x > 0 and 1 y s. Under the stability condition λ sµ < 1, we denote by p x,y the steady-state probability that the system is in state x, y). Thanks to the Markov chain structure, we solve the steady-state equations using standard results from the theory of linear difference equations see for example Queffélec and Zuily 2013)). For y = s and x > 0, we have p x,s λ+sµ 0 ) = λp x 1,s. Then p x,s = λ λ+sµ 0 ) x p0,s. For 0 y < s, and x > 0 we have λ + s y)µ + yµ 0 )p x,y = λp x 1,y + s y)µp x+1,y + y + 1)µ 0 p x+1,y+1. 9) The homogeneous equation associated to Equation 9) is s y)µz 2 λ + s y)µ + yµ 0 )z + λ = 0, 10) 11

12 with z as a variable, for z C. It has two solutions denoted by z y and z y and are given by z y = 1 λ + s y)µ + yµ 0 ) λ + s y)µ + yµ 0 ) 2s y)µ 2 4s y)λµ, 11) z y = 1 λ + s y)µ + yµ 0 + ) λ + s y)µ + yµ 0 ) 2s y)µ 2 4s y)λµ, 12) for 0 y < s. In Proposition 2, we provide the intervals where z y and z y are ranging. Proposition 2 For 0 y < s, we have 0 z y < 1 and z y > 1. Proof. Let us first prove that z y > 1. We have µ 0 > 0. Since z y increases in µ 0, Equation 12) implies z y > 1 λ + s y)µ + ) λ + s y)µ) 2s y)µ 2 4s y)λµ. 13) Observing that λ + s y)µ) 2 4s y)λµ = λ s y)µ) 2, Inequality 13) becomes z y > 1 λ + s y)µ + λ s y)µ ), 2s y)µ where t is the absolute value of t, for t R. Consider the case λ s y)µ, thus λ s y)µ = λ + s y)µ, which leads to z y > 1. Consider now the remaining case, i.e., λ > s y)µ. Then λ s y)µ = λ s y)µ, which implies z y > λ s y)µ > 1. In summary, we have z y > 1. Let us now prove that 0 z y < 1. From Equation 10), we may write s y)µz y z y = λ. Since λ 0 and z y > 1 > 0, we obtain z y 0. In what follows, we prove that 0 z y < 1. For λ = 0 we have z y = 0, then the result immediately follows. For λ > 0, the derivative of z y in µ 0 is given by λ ) z y + s y)µ + yµ0 ) 2 4s y)λµ λ + s y)µ + yµ y 0 ) = µ 0 2s y)µ, λ + s y)µ + yµ 0 ) 2 4s y)λµ for µ 0 > 0. It is straightforward to see that the numerator is strictly negative, for λ > 0, and the denominator is strictly positive. Therefore, z y is strictly decreasing in µ 0. Equation 11) then implies z y < 1 λ + s y)µ ) λ + s y)µ) 2s y)µ 2 4s y)λµ, 14) for µ 0 > 0. Using the same discussion as that after Inequality 13), we deduce that z y < 1. This 12

13 finishes the proof of the proposition. Because of the last term in the right hand side of Equation 9) and the fact that the 2s + 1) roots z 0, z 1,, z s, z 0, z 1,, z s are all distinct, p x,y can be written as a linear combination of z x i and z x i for y i s. Since z y > 1, the convergence of the stationary probabilities forces the linear factors of z x y to be all equal to zero. We therefore obtain, for 0 y s and x 0, with z s = λ λ+sµ 0 p x,y = s A i,y zi x, 15) i=y and A i,y R, for 0 y s and y i s. The stationary probabilities are now written as a function of a finite number of unknown parameters, namely the A i,y for 0 y s and y i s. In what follows, we compute these s+1)s+2) 2 parameters. Using Equation 9), we obtain s y)µzi 2 A i,y+1 = A + λ + s y)µ + yµ 0)z i λ i,y y + 1)µ 0 zi 2, 16) for 0 y < i s. Recall that z i is a root of the equation s i)µz 2 λ + s i)µ + iµ 0 )z + λ = 0. s i)µzi Thus, the quantity A 2+λ+s i)µ+iµ 0)z i λ i,y y+1)µ 0 zi 2 the right hand side of Equation 16) leads to for 0 y < i s. Therefore is equal to zero. Subtracting this quantity from A i,y+1 = A i,y i y)µ1 z i ) µ 0 ) y + 1)µ 0 z i, 17) i 1 A i,y = A i,i k=y = A i,i k + 1)µ 0 z i i k)µ1 z i ) µ 0 ) = A µ 0 z i i,i µ1 z i ) µ 0 ) µ 0 z i y ) i i, µ1 z i ) µ 0 y ) i y i! y!i y)! 18) for 0 y < i < s. Thus, it remains to compute the s + 1 parameters A y,y, for 0 y s. Using Equation 15), we obtain p 0,s = A s,s and p 0,s 1 = A s 1,s 1 + A s,s 1. From Equation 18), we may sµ 0 z s write A s,s 1 = A s,s µ1 z s ) µ 0. Using now the boundary equation λp 0,s = µp 0,s 1 and z s = implies the following relation between A s 1,s 1 and A s,s : λ λ+sµ 0 λ λ + sµ 0 A s 1,s 1 = A s,s µ λ + sµ 0 µ). 19) 13

14 The others boundary equations are p 0,y λ + s y)µ) = s y)µp 1,y + y + 1)µ 0 p 1,y+1 + s y + 1)µp 0,y 1, 20) for 0 < y < s. Using Equation 15), we have p 0,y = s A i,y, p 1,y = s A i,y z i and p 1,y+1 = s i=y+1 A i,y+1 z i. Then, using Equation 17) we obtain p 1,y+1 = s p 0,y λ + s y)µ) s y)µp 1,y y + 1)µ 0 p 1,y+1 = 0 < y < s. Moreover, using Equation 18) implies s i=y i=y i=y i y)µ1 z A i ) µ 0 ) i,y i=y+1 y+1)µ 0. Thus A i,y λ + s i)µ1 z i ) + i y)µ 0 ), for p 0,y λ + s y)µ) s y)µp 1,y y + 1)µ 0 p 1,y+1 21) s ) µ 0 z i y ) i i = A i,i λ + s i)µ1 z i ) + i y)µ 0 ), µ1 z i ) µ 0 y i=y for 0 < y < s. Finally we deduce from Equation 20) that A y 1,y 1 = s A i,i i=y µ 0 z i µ1 z i ) µ 0 ) i y ) i λ + s i)µ1 zi ) + i y)µ 0 y s y + 1)µ y µ 0 z i i y + 1 µ1 z i ) µ 0 22) ), for 0 < y < s. Note that this expression is also true for y = s; replacing y by s in Equation 22) leads to Equation 19). Since all probabilities sum up to one, we have s s y=0 i=y A i,i 1 z i µ 0 z i µ1 z i ) µ 0 ) i y ) i = 1. 23) y Equations 19), 22) and 23) form a system of s+1 independent linear equations, that can be easily numerically solved, and leads to the coefficients A i,i, for 0 i s. This finishes the characterization of all steady-state probabilities, p x,y, for x 0 and 0 y s. The throughput T λ, µ, µ 0, s) may be written as T λ, µ, µ 0, s) = µ 0 = µ 0 s s yp x,y 24) y=1x=0 s y=1 i=y ya i,y 1 z i. 14

15 From the stability condition on inbound calls, we may write or equivalently Since s y=0x=0 p x,y = 1 and s y=0x=0 λ = s s y)µp x,y, y=0x=0 s λ = µ s p x,y y=0x=0 s y=0x=0 yp x,y. yp x,y = T λ,µ,µ 0,s) µ 0 we obtain T λ, µ, µ 0, s) = µ 0 s λ µ ). Note that for µ 0 = µ, the result here coincides with that of the previous section, i.e., T λ, µ, µ, s) = sµ λ. As for the call waiting performance, it is given by P W > τ) = s p x,y P W > τ x, y)), y=0x=0 where P W > τ x, y)) is the conditional probability that the waiting time of a new call exceeds τ, given that it finds y s in service, s y calls in service, and x calls waiting ahead in the queue, for 0 y s and x 0. The computation of P W > τ x, y)), for 0 y s and x 0, is as follows. For x = 0 and 0 y s, the new call has to wait for a service completion of one of the y s, or one of the s y calls, so, P W > τ 0, y)) = e τyµ 0+s y)µ). For x = 1 and 0 < y s, the probability that the next service completion is that of an is yµ 0 yµ 0 +s y)µ. Thus, the waiting time of the new call follows a hypoexponential distribution consisting of the summation of two exponential random variables with rates yµ 0 + s y)µ and y 1)µ 0 + s y + 1)µ with probability yµ 0 yµ 0 +s y)µ, and it follows an Erlang distribution with 2 phases and yµ 0 + s y)µ as a rate per stage with probability 1 P W > τ 1, y)) = for 0 y s. yµ 0 yµ 0 + s y)µ yµ 0 yµ 0 +s y)µ. This leads to y 1)µ 0 + s y)µ)e τyµ 0+s y)µ) yµ 0 + s y)µ)e τy 1)µ 0+s y)µ) µ µ 0 s y)µ + yµ 0 + s y)µ e τyµ 0+s y+1)µ) 1 + τyµ 0 + s y)µ)), One can continue in the same way to derive all the conditional waiting time probabilities for x > 1, which finishes the characterization of the performance measures throughput and call waiting time distribution) in the case of unequal service rates. 15

16 3.2 Construction of the Adaptative Threshold Policy In this section, we use the performance evaluation results to find an insight on how we should adapt the threshold as a function of the intensity of the call arrivals. The objective is to maximize the throughput of s while reaching the constraint on the call waiting times for the whole day. We find that during the periods with low demand, the need of having a good service level is more important than during the periods with high demand. On the basis of this observation, we build a method for adapting the threshold. optimal threshold policy. We then evaluate this method by comparing it with the Numerical Observations For a given time interval long enough to reach the stationary regime, one can use the results of Section 3.1 to obtain the optimal threshold, denoted by u, for Problem 1). Consider now a working day with two time intervals, each with a different call arrival rate, and on each of which the stationary regime is reached. Following common practice and most call center models in the literature, it is appropriate to assume that a system with constant parameters achieves a steadystate quickly within short - half hour or hour - intervals Green et al. 2001); Gans et al. 2003)). We want to find the optimal couple of thresholds that answers our optimization problem, where the call service level constraint is for the whole day. We denote the proportion of the length of the first second) time interval by I 1 I 2 ) and the corresponding mean arrival rate by λ 1 λ 2 ). We have I 1 + I 2 = 1. Without loss of generality, we consider cases where λ 1 λ 2. In Table 1, we consider various scenarios for arrival rates, service rates, and relative time durations between the two intervals. Using the results of Section 3.1, we give the optimal threshold of each interval in isolation, i.e., the highest threshold which verifies the service level constraint. They are denoted by u 1 and u 2 for I 1 and I 2, respectively. The symbol in Table 1 is used for the cases where the call service level can not be met, even with a threshold equal to zero. We also evaluate the couple of thresholds which answers Problem 1) on the set of the two intervals. This couple is found by an exhaustive test of all the possible values for the couple u 1,u 2 ). We denote by u 1, u 2 ) this optimal couple. Remark that for this couple, Problem 1) does not have to be answered on each interval but on the set of the two intervals. Finally, we give the performance measures for each interval and for the set of the two intervals for the couple u 1, u 2 ). In summary, our optimization problem can be formulated as finding the best couple u 1, u 2 ) that answers the following problem { Maximize I 1 T λ1 u 1 ) + I 2 T λ2 u 2 ) subject to I 1 λ 1 λ 1 +λ 2 SL λ1 u 1 ) + I 2 λ 2 λ 1 +λ 2 SL λ2 u 2 ) α. 25) 16

17 Table 1: Optimal couples of thresholds s = 10, τ = 30 seconds, α = 80%) λ 1 λ 2 µ µ 0 I 1 I 2 u 1 u 2 u 1, u 2) P W 1 < τ) P W 2 < τ) P W < τ) T 1 T 2 T % 50% 8 8 8,8) 84.04% 84.04% 84.04% % 50% 8 6 8,7) 84.04% 77.99% 80.62% % 50% 9 8,4) 96.81% 74.79% 80.30% % 33% 8 6 8,7) 84.04% 77.99% 81.66% % 20% 8 6 8,8) 84.04% 69.15% 80.39% % 10% 9 9,7) 88.19% 63.94% 82.13% % 50% 8 7,5) 90.92% 72.93% 80.13% % 50% 10 10,7) 89.34% 74.94% 80.70% % 20% 10 10,10) 89.34% 67.56% 83.40% % 50% 9 8 9,9) 83.51% 77.09% 80.18% % 20% 9 8 9,10) 83.51% 68.19% 80.26% % 20% 9 9 9,10) 88.63% 60.45% 82.10% % 50% 9 9 9,9) 88.63% 87.77% 88.18% % 50% 9 8 9,8) 88.19% 84.04% 85.42% % 50% ,7) 59.34% 90.92% 85.66% % 50% ,8) 61.33% 84.04% 81.97% % 50% ,8) 63.03% 84.04% 83.83% Table 2: Value of a/s above which SLu) decreases in a s = 10, τ = 30 seconds, µ = µ 0 = 0.2) u a/s An important observation from Table 1 is that the choices for the threshold are done for values of u close to s. The reason is related to the concavity of the call service level see Proposition 1). We observe three possible situations corresponding to the three parts of Table 1. In the first situation u 1 respectively u 2 ) is always higher or equal to u 1 respectively lower or equal to u 2 ) for the optimal couple u 1, u 2 ). In the second one we have u 1, u 2 ) = u 1, u 2 ). In the last situation u 1 respectively u 2 ) is always lower or equal to u 1 respectively higher or equal to u 2 ) for the optimal couple u 1, u 2 ). Although the first situation does not seem to be the most intuitive one, it corresponds to most cases. In order to respect the overall call service level, we observe that we should strictly respect the service level during the interval with a small arrival rate I 1 ), and more flexibility is accepted when the arrival rate is high I 2 ). In what follows, we explain why this insight holds in most practical cases. We first justify that SLu) SLu) = SLu + 1) SLu) for 0 u < s) is decreasing in the workload in most practical cases. Second, using this assumption we prove that there is less waste for the call service level, when increasing the threshold during higher workload periods. Finally, we derive the required conditions under which the insight does hold. Figure 2 reveals that, as the workload increases, the sensitivity of the service level for a given threshold SLu) = SLu + 1) SLu) for 0 u < s) first increases and then decreases. In 17

18 ddl dl dl ddl ddl edl edl dddl ^> d ^> e Figure 2: Sensitivity of the service Level s = 10, τ = 30 seconds, µ = µ 0 = 0.2) Lemma 1, we prove for µ = µ 0 that the last part of the curves SLu) decreases in the workload. Table 2 provides some numerical illustrations for the value of a/s above which the curve SLu) decreases in a. We observe for this example that the values of a/s are lower than 80%. In practice the agent utilization in call centers is usually higher than 80% see Koole 2013)). If a situation with a low workload happens, the threshold would increase and reach its maximal values u = s 1 or u = s). Since the last part of the curves SLu) as function of the workload decreases, the practical situations are likely to be those where the sensitivity of SLu) decreases in the workload. Lemma 1 The following holds for µ = µ 0, 0 a s and s 2. SLs 1) is decreasing in a if s 1 τµ 0, otherwise SLs 1) is first increasing then decreasing in a. There exists a value of a, 0 < a < s, above which SLu) is decreasing in a, for 0 u < s 1. Proof. Using Theorem 1, we have SLu) = SLu + 1) SLu) = e τsµ1 a/s) Cs, u + 1, a) Cs, u, a), for 0 u s 1. Let us now prove the first statement of the Lemma. Replacing u by 18

19 s 1 in Equation 3) leads to Cs, s 1, a) = as s 1) s 1)! s!1 a/s) = s s 1) a/s 1 + a/s + a/s) a/s) 1 a/s = a/s, a k s 1)! s 1) + k)! + as s 1) s 1)! a s! s a ) 1 1 for 0 a s. We also have Cs, s, a) = 1. Thus SLs 1) = e τsµ1 a/s) 1 a/s), for 0 a s. We have SLs 1) a = 1 s e τsµ1 a/s) 1 + s a)τµ), for 0 a s. The expression 1 + s a)τµ decreases in a. The equation, 1 + s a)τµ = 0, in the variable a is equivalent to a = s 1 τµ. If s 1 SLs 1) τµ 0 then a 0 for 0 a s and SLs 1) decreases in a. Otherwise if s 1 τµ s 1 τµ and then decreasing from s 1 τµ > 0, SLs 1) is first increasing from 0 to to s as a function of a 0 a s). Note that s 1 τµ < s. We then deduce that the last part of the curve of SLs 1) is always decreasing as a function of a. We next prove the second statement. We can write Cs, u, a) as Cs, u, a) = as u u! s!1 a/s) p u. If 0 u < s, we have lim a 0, a > 0 a s u u! s!1 a/s) = 0. From the proof of Proposition 1, we know that lim a 0, a > 0 p u = 1. We then obtain lim a 0, a > 0 Cs, u, a) = 0, for 0 u < s. We can also write Cs, u, a) as Cs, u, a) = a/s + s!1 a/s) s u ) 1 a k+u s u+k)!. We 19

20 have lim a s, a < s s u a k+u s a/s + s!1 a/s) u + k)! = 1, for 0 u s. Thus lim a s, a < s Cs, u, a) = 1, for 0 u s. Since we have e τsµ e τsµ1 a/s) 1 for 0 a s, we obtain lim a s, a < s SLu) = lim a 0, a > 0 SLu) = 0, for 0 u < s 1. Since SLu) is not zero, the curve of SLu) has at least one extremum in the variable a for 0 < a < s. This proves that there exists a value of a 0 < a < s) after which SLu) decreases in a for 0 u < s 1, and finishes the proof of the lemma. Assuming now that the sensitivity of the call service level is decreasing in the workload, we prove in Proposition 3 that there is less waste for the call service level, when increasing the threshold during higher workload periods. For the call service level constraint in Problem 25), Corollary 1 completes Proposition 3 by providing the necessary conditions under which it is better to increase the threshold during high workload periods. Proposition 3 If λ 1 < λ 2 and SLu) is decreasing in the workload, then SL λ1 u 1 ) > SL λ 2 u 2 ). Proof. We distinguish two cases; u 1 = u 2 or u 1 > u 2. The other case u 1 < u 2 does not exist because λ 1 < λ 2. If u 1 = u 2 then increasing u is less sensitive in SL λ 2 than in SL λ1 since the sensitivity of SL is decreasing in the workload, i.e., SL λ1 u 1 ) > SL λ 2 u 2 ). Consider now the case u 1 > u 2. Since SL is decreasing and concave in u see Proposition 1), we deduce that SL λ1 is more sensitive to the increasing of u starting from u 1 than starting from u 2, i.e., SL λ1 u 1 ) > SL λ 1 u 2 ). Starting from u 2, SL λ 1 SL λ2, i.e., SL λ1 u 2 ) > SL λ 2 u 2 ). As a consequence SL λ 2 is more sensitive to the increasing of u than is less sensitive to the increasing of u starting from u 2 than SL λ 1 would be starting from u 1, i.e., SL λ 1 u 1 ) > SL λ 2 u 2 ). Corollary 1 If I 2 SL λ2 u 2 ) I 1 SL λ1 u 1 ) λ 2 < λ 1 < λ 2 and SLu) is decreasing in the workload, there is less waste for the call service level on the two intervals, when increasing the threshold during the higher 20

21 Table 3: Interval of validity of Corollary 1 λ 2 = 1.3, s = 10, µ 0 = µ = 0.2, τ = 30s, α = 80%) λ SL λ2 u 2 ) SL λ1 u ) λ SL λ2 u 2 ) SL λ1 u ) λ 2 < λ 1 1 False False True True True True True True workload period. Proof. Using Proposition 3 we know that SL λ1 u 1 ) > SL λ 2 u 2 ), or SL λ 2 u 2 ) SL λ1 u 1 ) < 1. We can then find values of λ 1, λ 2, I 1 and I 2 which satisfy the inequality I 2 SL λ2 u 2 ) I 1 SL λ1 u ) λ 2 < λ 1 < λ 2, then 1 with the first inequality we have I 2 SL λ2 u 2 ) λ 2 < I 1 SL λ1 u 1 ) λ 1, and finally I 1 λ 1 λ 1 + λ 2 SL λ1 u 1) > I 2 λ 2 λ 1 + λ 2 SL λ2 u 2). This finishes the proof of the corollary. In practice the changes in the threshold are likely to be made by the ACD at predefined and equal intervals of time, i.e., I 1 = I 2. The condition, for the result in Corollary 1 to hold, then becomes SL λ2 u 2 ) SL λ1 u 1 ) < λ 1 λ 2 < 1. Note that this condition does not happen only for the extreme situations with very high differences between the mean arrival rate values λ 1 << λ 2 ). An illustration is given in Table Our Adaptive Threshold Policy ATP) We propose for Problem 1) an adaptive threshold policy which adjusts the threshold as a function of the call workload. This policy is based on the the first and second order monotonicity properties of the performance measures as a function of the threshold u, and on the observation drawn in Section As mentioned in Section 2, the threshold is reevaluated at the beginning of each interval i i = 1,..., N). The threshold associated to interval i is denoted by u i. The global service level for the whole day all N intervals) is denoted by SL, and the global one from interval 1 to interval i is denoted by SL i, for i = 1,..., N. If SL i is higher lower) than α at the beginning of an interval i i = 2,..., N) then the policy increases decreases) the threshold. To update the threshold, we use a real parameter denoted by c i i = 1,..., N). The threshold u i is defined as the closest integer to c i, for i = 1,..., N. Note that the parameter c i is chosen to be real in order to smooth the change in the threshold u i. We start with u 1 = c 1 = s. For i 2, if we need to increase the threshold in case if SL i > α), then 21

22 we consider c i = c i c i 1 /s. If we need to decrease the threshold in case SL i < α), then c i = c i 1 c i 1 /s. In the remaining case SL i = α), we consider c i = c i 1. This policy is refereed to as ATP. In what follows, we discuss the efficiency of how ATP updates the threshold. The main two characteristics of ATP are: An Increasing decreasing) of the threshold in case the measured call service level is better worse) than the target service level, A decreasing speed in the increasing decreasing) of the threshold when this threshold increases decreases). From Proposition 1, we know that the throughput increases and the call service level decreases in u. Thus, the threshold should be increased when the measured service level is better than the target service level, and vice versa. This justifies the first characteristic of ATP. d d d ed d d d ed W tm d dd d d e e dd d d d ed d d d ed W tm d dd d d e e dd a) λ = 0.1 b) λ = 0.5 d d d ed d d d ed W tm d dd d d e e dd d d d ed d d d ed W tm d dd d d e e dd c) λ = 1 d) λ = 1.4 Figure 3: Performance measures s = 10, µ = µ 0 = 0.2, τ = 0.5, α = 80%) The second characteristic of ATP is justified by the convexity of the performance measures and the correlation between them. Consider a situation with sufficiently high call service levels, for example during light workload periods. The threshold u should then reach high values close to the number of agents. For high values of the threshold, we know from Proposition 1 that the 22

23 call service level is decreasing and concave, and the throughput is increasing and concave in u. An illustration is given in Figures 3a) and 3b). Therefore, increasing u would go with only a little improvement in the throughput, and at the same time a high loss in the call service level. This situation is well managed by ATP. As u increases, ATP decreases the speed of increasing u, which reduces the non efficient situations with high values of the threshold. Moreover, ATP behaves as required by the insight derived in Section From the insight, we know that for the optimization problem 1) we should strictly respect the call service level constraint during light workload periods. ATP conservatively increases a high threshold, which is the way to give importance to the respect the call service level constraint. Consider now a situation with bad call service levels, for example during high workload periods. The threshold u should then reach small values. For small values of the threshold, we know from Proposition 1 that the call service level is decreasing and concave in u, i.e., almost not sensitive to the decreasing of u. This does not hold for the throughput. An illustration is given in Figures 3c) and 3d). Therefore, decreasing u would go with only a little improvement in the call service level, and at the same time a high loss in the throughput. This situation is again well managed by ATP. As u decreases, ATP decreases the speed of decreasing u, which reduces the non efficient situations with small values of the threshold. Again, ATP behaves as required by the insight derived in Section From the insight, we know that for the optimization problem 1) it is tolerated to violate the call service level constraint during high workload periods. ATP conservatively decreases a low threshold, which is the way to give less importance to the respect of the call service level constraint Evaluation of the Adaptative Threshold Policy In this section, we evaluate the quality of the ATP policy by comparing it with the optimal one. First we provide the optimal threshold policy. Because of the discrete nature of the threshold, one may see that the threshold should vary between two or more values. The reason is that we need to exactly satisfy the constraint on calls in Problem 1) in order to maximize the throughput. From Bhulai and Koole 2003), we know that to exactly satisfy the constraint on calls, randomization is optimal for threshold policies. For both cases µ 0 = µ and µ 0 µ, Theorem 2 provides a weak condition that leads to the optimal randomization policy between two threshold values. A randomized threshold policy, between two thresholds u 1 and u 2 and with a randomization parameter p [0, 1], works as follows. At each event an inbound call arrival or a service completion), the value of the threshold value changes from u 1 to u 2 with probability p, stays in u 1 with probability 1 p, changes from u 2 to u 1 with probability 1 p, stays in u 2 with probability p. 23

24 Theorem 2 Consider 0 u 1, u 2 s such that SLu 1 ) α SLu 2 ). If there exists γ R for which randomizing between u 1 and u 2 maximizes T u) + γslu) and leads to a call service level exactly equal to α, then randomizing between u 1 and u 2 is optimal. Proof. Let p [0, 1] be the parameter of randomization between u 1 and u 2. Assume that we can find a couple u 3, u 4 ) u 1, u 2 ) and a parameter of randomization q [0, 1] such that the constraint on calls is also saturated and SLu 3 ) α SLu 4 ). We have pt u 1 ) + 1 p)t u 2 ) + γpslu 1 )+γ1 p)slu 2 ) qt u 3 )+1 q)t u 4 )+γqslu 3 )+γ1 q)slu 4 ). Since γpslu 1 )+ γ1 p)slu 2 ) = γqslu 3 ) + γ1 q)slu 4 ) = γα, we deduce that pt u 1 ) + 1 p)t u 2 ) qt u 3 ) + 1 q)t u 4 ). Then the couple u 1, u 2 ) is optimal, which completes the proof. The randomization between two thresholds allows for the constraint on calls to be met exactly. For our system with constant parameters, we believe that the randomization is between two successive thresholds. Since the throughput is neither convex nor concave it is difficult to rigorously prove this result. However, if we denote by u 0 u s) the highest threshold that verifies SLu ) > α, we numerically checked that with γ = T u +1) T u ) SLu +1) SLu ) for 0 u < s), the expression T u)+γ SLu) is strictly increasing from u = 0 to u = u, strictly decreasing from u = u +1 to u = s and T u ) + γslu ) = T u + 1) + γslu + 1). Then for all the considered numerical situations the optimal policy is a randomization between two adjacent values when 0 u < s. When u = s, the optimal policy is to keep the threshold constant and equal to s. In Table 4, we propose 5 representative scenarios with constant arrival rates and compare the optimal throughput with the one found with under ATP. Although the ATP method is not optimal, the difference with the optimum is quite small. This shows the advantage of ATP in the case of constant arrival rates. Recall that our main purpose in this paper is the analysis of the case with a fluctuating arrival rate. In the next section, we consider the case of a fluctuating arrival rate and evaluate the performance of ATP through a comparison with other intuitive methods. 4 Non-Constant Arrival Rates In Section 4.1 we compare ATP with methods that use constant step sizes. Then in Section 4.2 we analyze the impact of the parameters on the choice of the method. In Section 4.3 we propose some other intuitive adaptive methods. We consider cases where the length of the working day equals eight hours D = 8h) and a frequent possibility of reevaluating the real threshold c, at the beginning of each time interval with length θ = 1 min, 5 min or 15 min. We use simulation to obtain the performance measures. For each 24